Newton’s laws of motion
(A)first law of motion and its application
A body will continue in its present state of rest or, if it is in motion, will continue to move with uniform speed in a straight line unless it is acted upon by a force. This tendency of a body to remain in its state of rest or uniform motion is called the inertia of the body. For this reason, Newton’s first law is sometimes called ‘the law of inertia’.
Note: momentum is an important property of a moving object. It explains its tendency to continue moving in a straight line.
The momentum of a body is defined as the product of its mass and its velocity.
Momentum = mv.
(B) second law of motion and its application
The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.
Force = change in momentum/ time
Suppose a force f acts on a body of mass m for a time t and causes it to change its Velocity from u to v, then Newton’s second law can stated as:
F = mv-mu/t
So f = m(v-u)/t
But v-u/t = a(acceleration)
F = ma, f= kma
Note: the SI unit of force is called the Newton (N). Thus when f is in Newtons, m in kg, a in ms-2 we have f = ma.
Deduction from Newton’s second law
Impulses of a force
Now f = m(v-u)/t
Or. Ft = mv – my
The quantity ft. is called impulse I of the force. Impulse is the product of a large force and a very short time during which it acts . The unit of impulse is the Newton second (Ns). Since the quality is equal to change in momentum it means that momentum can also be expressed in Ns. I = ft. = change in momentum
Note: thus for a body of mass m falling under gravity alone, the force of gravity, f, acting on it is given by f = ma
Meaning f = mg (measured in Newtons)this force is called the weight W of the body. Hence we write w=mg.
Note: weight is a force and should be expressed in Newtons (N).
Weight of a body in a lift (weightlessness) A lift is a device operated electrically and used for moving people and load up or down a tall building.
- When the lift is stationary or moving with a constant Velocity we have that w = mg = R. Where m is the mass of the man.
- When the lift accelerates upwards with an acceleration a, the man is pulled upwards with an acceleration a. The unbalanced force (f) on him is given by: f = R – mg = ma hence R = m(a+g). Apparent weight of the man when the lift accelerates upwards is given by w = R = m(a+g). The man appears to weigh less under this condition
- When the lift descends with an acceleration a= g, (free fall). Thus the man’s apparent weight is equal to zero, i.e. he appears to have no weight. Such a situation is referred to as weightlessness.
(C)third law of motion and its application
To every action there is an equal and opposite reaction.
Note: since force is proportional to change in momentum, it follows that the momentum of the bullet is equal and opposite to the momentum of the gun. Thus for a bullet of mass, m and velocity, v, the Velocity V of recoil of the gun is given as MV = mv
V = -mv/M
Where M is the mass of the gun.
The rocket is a good example of Newton’s third law of motion
Law of conservation of momentum
Newton’s second and third laws enable us to formulate an important conservation law known as the law of conservation of momentum. Which states
In a system of colliding objects the total momentum is conserved, provided there is no net external force acting on the system. It can also be stated as follows: the total momentum of an isolated or closed system of colliding bodies remains constant. Thus if two or more bodies collide in a closed system, the total momentum before collision is equal to the total momentum after collision. By a closed or isolated system we mean that system on which no external forces act. Let u1, u2 and v1, v2 be the initial and final velocities of the two colliding bodies of masses m1 and m2. The conservation law can be stated as:
m1u1 + m2u2 = m1v1 + m2v2
Note: all the velocities must be measured in the same direction along the same lines, with correct positive or negative signs.
Conservation of momentum
The argument applies to bodies which are elastic and rebound from each other after collision with different Velocities. It also applies to bodies which are inelastic and join together after the collision and move away with the same Velocity. In this case.v1 = v2 = v
m1u1 + m2u2 = (m1 + m2)v
- An object is situated within the Earth’s gravitational field. Which of the following factors does not affect the acceleration of free fall g? (wassce 2000) A. The distance of the object from the center of the earth .B. The latitude of the earth on which the object is situated C. The mass of the object D. The rotation of the earth
Solution: this because when falling the force of gravity pulls the mass of an object downward which makes it weightlessness.
2. State Newton’s second law of motion(wassce 2000 theory)
Answer: the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.
3. Show that f=ma where F is the magnitude of the force acting on a body of mass m to give it an acceleration of magnitude a.(Wassce 2000 theory)
Answer: force = change in momentum/time
Suppose a force f acts on a body of mass m for a time t and causes it to changes its Velocity from u to v, then Newton’s second law can be stated as:
So f = m(v-u)/t
But. v-u/t = a(acceleration)
F = ma, f = kma
4. The tendency of a body to remain at rest when a force is applied to it is called (Wassce 2003)
Solution: inertia can also be applied when it comes to the first law of motion.
5. The force with which an object is attracted to the earth is called its( jamb 1983)
Solution: gravity pulled the weight of an object to the center of the earth
6. A force of 100N is used to kick a football of mass 0.8kg. Find the Velocity with which the ball moves if it takes 0.8s to be kicked.(jamb 2003)
Solution: f= m(v-u)/t; 100= 0.8(v-0)/0.8
v = 100m/s
7. I jet-propelled aircraft
ll. rocket propulsion
III. the recoil of a gun
IV. a person walking
Which of the above is based on Newton’s third law of motion?(jamb 2004)
A. I, ii, III and IV
B. I and III only
C. I and ii only
D. I, ii and III only
Solution: Ans B dwells on the concept that says to every action there is an equal and opposite reaction which is Newton’s third law of motion.
8. Calculate the apparent weight loss of a man weighing 70kg in an elevator moving downwards with an acceleration of 1.5ms-2(jamb 2013)
Solution: wt loss = mg-m(g-a)
= 70 × 10 – 70(10 – 1.5)=105N